142787
A spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the surface tension of the liquid, then the diameter of the body is
1 $\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$
2 $\sqrt{\frac{6 \sigma}{g(2 \rho-d)}}$
3 $\sqrt{\frac{4 \sigma}{g(2 \rho-d)}}$
4 $\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
Explanation:
D Balancing the surface tension of the drop $=$ buoyant force + force of surface tension $\Rightarrow \frac{4}{3} \pi \mathrm{r}^{3} \rho g=2 \pi \mathrm{r} \sigma+\frac{1}{2}\left(\frac{4}{3} \pi \mathrm{r}^{3} \mathrm{dg}\right)$ $\Rightarrow 2 \pi \mathrm{r} \sigma=\frac{\pi \mathrm{r}^{3} \mathrm{~g}}{3}[4 \rho-2 \mathrm{~d}]$ $\Rightarrow r^{2}=\frac{3 \times 2 \pi \sigma}{\pi g(4 \rho-2 d)}$ $\Rightarrow r^{2}=\frac{3 \sigma}{g(2 \rho-d)}$ $\Rightarrow r=\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$ $\text { Diameter }=2 r=\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
AP EAMCET (22.04.2018) Shift-II
Mechanical Properties of Fluids
142788
Two circular plates of radius $5 \mathrm{~cm}$ each, have a $0.01 \mathrm{~mm}$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $=\mathbf{7 3} \mathrm{dyne} / \mathrm{cm})$ ?
1 $125 \mathrm{~N}$
2 $95 \mathrm{~N}$
3 $115 \mathrm{~N}$
4 $105 \mathrm{~N}$
Explanation:
C Given, surface tension of water $=73$ dyne $/ \mathrm{cm}$ We know that, $\mathrm{W}=\mathrm{F} \times \mathrm{d}$ $\mathrm{W}=\mathrm{T} \times \mathrm{A} \times 2$ by using equation (i) and (ii), we get $\mathrm{F} \times \mathrm{d}=\mathrm{T} \times \mathrm{A} \times 2$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \mathrm{A}}{\mathrm{d}}$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \pi \mathrm{r}^{2}}{\mathrm{~d}}$ $\mathrm{F}=\frac{2 \times 73 \times 10^{-3} \times \pi \times(0.05)^{2}}{0.01 \times 10^{-3}}$ $\mathrm{F}=36.5 \pi \simeq 115$ Newton
BITSAT-2010
Mechanical Properties of Fluids
142789
If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be
1 $1: 4$
2 $4: 1$
3 $2: 1$
4 $1: 2$
Explanation:
D According to the kinetic energy gas equation $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mNv}^{2}}{\mathrm{~V}}$ $\therefore \quad \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ and $\quad P_{f}=\frac{N\left(\frac{1}{2} m\right)(2 v)^{2}}{3 V}=\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ $\left\{\right.$ Given, $\left.\mathrm{v}_{\mathrm{f}}=2 \mathrm{v}, \mathrm{m}_{\mathrm{f}}=\mathrm{m} / 2\right\}$ Ratio of pressure from initial to final $\therefore \quad \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{f}}}=\frac{\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}}{\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}}=\frac{1}{2}$ Hence, ratio of initial and final pressure is $1: 2$.
CG PET 2019
Mechanical Properties of Fluids
142790
Drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\rho_{0}$. If surface tension of liquid is $\mathrm{s}$, the radius of drop is
142787
A spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the surface tension of the liquid, then the diameter of the body is
1 $\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$
2 $\sqrt{\frac{6 \sigma}{g(2 \rho-d)}}$
3 $\sqrt{\frac{4 \sigma}{g(2 \rho-d)}}$
4 $\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
Explanation:
D Balancing the surface tension of the drop $=$ buoyant force + force of surface tension $\Rightarrow \frac{4}{3} \pi \mathrm{r}^{3} \rho g=2 \pi \mathrm{r} \sigma+\frac{1}{2}\left(\frac{4}{3} \pi \mathrm{r}^{3} \mathrm{dg}\right)$ $\Rightarrow 2 \pi \mathrm{r} \sigma=\frac{\pi \mathrm{r}^{3} \mathrm{~g}}{3}[4 \rho-2 \mathrm{~d}]$ $\Rightarrow r^{2}=\frac{3 \times 2 \pi \sigma}{\pi g(4 \rho-2 d)}$ $\Rightarrow r^{2}=\frac{3 \sigma}{g(2 \rho-d)}$ $\Rightarrow r=\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$ $\text { Diameter }=2 r=\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
AP EAMCET (22.04.2018) Shift-II
Mechanical Properties of Fluids
142788
Two circular plates of radius $5 \mathrm{~cm}$ each, have a $0.01 \mathrm{~mm}$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $=\mathbf{7 3} \mathrm{dyne} / \mathrm{cm})$ ?
1 $125 \mathrm{~N}$
2 $95 \mathrm{~N}$
3 $115 \mathrm{~N}$
4 $105 \mathrm{~N}$
Explanation:
C Given, surface tension of water $=73$ dyne $/ \mathrm{cm}$ We know that, $\mathrm{W}=\mathrm{F} \times \mathrm{d}$ $\mathrm{W}=\mathrm{T} \times \mathrm{A} \times 2$ by using equation (i) and (ii), we get $\mathrm{F} \times \mathrm{d}=\mathrm{T} \times \mathrm{A} \times 2$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \mathrm{A}}{\mathrm{d}}$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \pi \mathrm{r}^{2}}{\mathrm{~d}}$ $\mathrm{F}=\frac{2 \times 73 \times 10^{-3} \times \pi \times(0.05)^{2}}{0.01 \times 10^{-3}}$ $\mathrm{F}=36.5 \pi \simeq 115$ Newton
BITSAT-2010
Mechanical Properties of Fluids
142789
If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be
1 $1: 4$
2 $4: 1$
3 $2: 1$
4 $1: 2$
Explanation:
D According to the kinetic energy gas equation $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mNv}^{2}}{\mathrm{~V}}$ $\therefore \quad \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ and $\quad P_{f}=\frac{N\left(\frac{1}{2} m\right)(2 v)^{2}}{3 V}=\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ $\left\{\right.$ Given, $\left.\mathrm{v}_{\mathrm{f}}=2 \mathrm{v}, \mathrm{m}_{\mathrm{f}}=\mathrm{m} / 2\right\}$ Ratio of pressure from initial to final $\therefore \quad \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{f}}}=\frac{\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}}{\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}}=\frac{1}{2}$ Hence, ratio of initial and final pressure is $1: 2$.
CG PET 2019
Mechanical Properties of Fluids
142790
Drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\rho_{0}$. If surface tension of liquid is $\mathrm{s}$, the radius of drop is
142787
A spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the surface tension of the liquid, then the diameter of the body is
1 $\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$
2 $\sqrt{\frac{6 \sigma}{g(2 \rho-d)}}$
3 $\sqrt{\frac{4 \sigma}{g(2 \rho-d)}}$
4 $\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
Explanation:
D Balancing the surface tension of the drop $=$ buoyant force + force of surface tension $\Rightarrow \frac{4}{3} \pi \mathrm{r}^{3} \rho g=2 \pi \mathrm{r} \sigma+\frac{1}{2}\left(\frac{4}{3} \pi \mathrm{r}^{3} \mathrm{dg}\right)$ $\Rightarrow 2 \pi \mathrm{r} \sigma=\frac{\pi \mathrm{r}^{3} \mathrm{~g}}{3}[4 \rho-2 \mathrm{~d}]$ $\Rightarrow r^{2}=\frac{3 \times 2 \pi \sigma}{\pi g(4 \rho-2 d)}$ $\Rightarrow r^{2}=\frac{3 \sigma}{g(2 \rho-d)}$ $\Rightarrow r=\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$ $\text { Diameter }=2 r=\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
AP EAMCET (22.04.2018) Shift-II
Mechanical Properties of Fluids
142788
Two circular plates of radius $5 \mathrm{~cm}$ each, have a $0.01 \mathrm{~mm}$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $=\mathbf{7 3} \mathrm{dyne} / \mathrm{cm})$ ?
1 $125 \mathrm{~N}$
2 $95 \mathrm{~N}$
3 $115 \mathrm{~N}$
4 $105 \mathrm{~N}$
Explanation:
C Given, surface tension of water $=73$ dyne $/ \mathrm{cm}$ We know that, $\mathrm{W}=\mathrm{F} \times \mathrm{d}$ $\mathrm{W}=\mathrm{T} \times \mathrm{A} \times 2$ by using equation (i) and (ii), we get $\mathrm{F} \times \mathrm{d}=\mathrm{T} \times \mathrm{A} \times 2$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \mathrm{A}}{\mathrm{d}}$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \pi \mathrm{r}^{2}}{\mathrm{~d}}$ $\mathrm{F}=\frac{2 \times 73 \times 10^{-3} \times \pi \times(0.05)^{2}}{0.01 \times 10^{-3}}$ $\mathrm{F}=36.5 \pi \simeq 115$ Newton
BITSAT-2010
Mechanical Properties of Fluids
142789
If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be
1 $1: 4$
2 $4: 1$
3 $2: 1$
4 $1: 2$
Explanation:
D According to the kinetic energy gas equation $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mNv}^{2}}{\mathrm{~V}}$ $\therefore \quad \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ and $\quad P_{f}=\frac{N\left(\frac{1}{2} m\right)(2 v)^{2}}{3 V}=\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ $\left\{\right.$ Given, $\left.\mathrm{v}_{\mathrm{f}}=2 \mathrm{v}, \mathrm{m}_{\mathrm{f}}=\mathrm{m} / 2\right\}$ Ratio of pressure from initial to final $\therefore \quad \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{f}}}=\frac{\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}}{\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}}=\frac{1}{2}$ Hence, ratio of initial and final pressure is $1: 2$.
CG PET 2019
Mechanical Properties of Fluids
142790
Drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\rho_{0}$. If surface tension of liquid is $\mathrm{s}$, the radius of drop is
142787
A spherical body of density $\rho$ is floating half immersed in a liquid of density $d$. If $\sigma$ is the surface tension of the liquid, then the diameter of the body is
1 $\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$
2 $\sqrt{\frac{6 \sigma}{g(2 \rho-d)}}$
3 $\sqrt{\frac{4 \sigma}{g(2 \rho-d)}}$
4 $\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
Explanation:
D Balancing the surface tension of the drop $=$ buoyant force + force of surface tension $\Rightarrow \frac{4}{3} \pi \mathrm{r}^{3} \rho g=2 \pi \mathrm{r} \sigma+\frac{1}{2}\left(\frac{4}{3} \pi \mathrm{r}^{3} \mathrm{dg}\right)$ $\Rightarrow 2 \pi \mathrm{r} \sigma=\frac{\pi \mathrm{r}^{3} \mathrm{~g}}{3}[4 \rho-2 \mathrm{~d}]$ $\Rightarrow r^{2}=\frac{3 \times 2 \pi \sigma}{\pi g(4 \rho-2 d)}$ $\Rightarrow r^{2}=\frac{3 \sigma}{g(2 \rho-d)}$ $\Rightarrow r=\sqrt{\frac{3 \sigma}{g(2 \rho-d)}}$ $\text { Diameter }=2 r=\sqrt{\frac{12 \sigma}{g(2 \rho-d)}}$
AP EAMCET (22.04.2018) Shift-II
Mechanical Properties of Fluids
142788
Two circular plates of radius $5 \mathrm{~cm}$ each, have a $0.01 \mathrm{~mm}$ thick water film between them. Then what will be the force required to separate these plate $($ S.T. of water $=\mathbf{7 3} \mathrm{dyne} / \mathrm{cm})$ ?
1 $125 \mathrm{~N}$
2 $95 \mathrm{~N}$
3 $115 \mathrm{~N}$
4 $105 \mathrm{~N}$
Explanation:
C Given, surface tension of water $=73$ dyne $/ \mathrm{cm}$ We know that, $\mathrm{W}=\mathrm{F} \times \mathrm{d}$ $\mathrm{W}=\mathrm{T} \times \mathrm{A} \times 2$ by using equation (i) and (ii), we get $\mathrm{F} \times \mathrm{d}=\mathrm{T} \times \mathrm{A} \times 2$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \mathrm{A}}{\mathrm{d}}$ $\mathrm{F}=\frac{2 \times \mathrm{T} \times \pi \mathrm{r}^{2}}{\mathrm{~d}}$ $\mathrm{F}=\frac{2 \times 73 \times 10^{-3} \times \pi \times(0.05)^{2}}{0.01 \times 10^{-3}}$ $\mathrm{F}=36.5 \pi \simeq 115$ Newton
BITSAT-2010
Mechanical Properties of Fluids
142789
If the masses of all molecules of a gas are halved and their speeds doubled, then the ratio of initial and final pressure will be
1 $1: 4$
2 $4: 1$
3 $2: 1$
4 $1: 2$
Explanation:
D According to the kinetic energy gas equation $\mathrm{P}=\frac{1}{3} \frac{\mathrm{mNv}^{2}}{\mathrm{~V}}$ $\therefore \quad \mathrm{P}_{\mathrm{i}}=\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ and $\quad P_{f}=\frac{N\left(\frac{1}{2} m\right)(2 v)^{2}}{3 V}=\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}$ $\left\{\right.$ Given, $\left.\mathrm{v}_{\mathrm{f}}=2 \mathrm{v}, \mathrm{m}_{\mathrm{f}}=\mathrm{m} / 2\right\}$ Ratio of pressure from initial to final $\therefore \quad \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{f}}}=\frac{\frac{\mathrm{Nmv}^{2}}{3 \mathrm{~V}}}{\frac{2 \mathrm{Nmv}^{2}}{3 \mathrm{~V}}}=\frac{1}{2}$ Hence, ratio of initial and final pressure is $1: 2$.
CG PET 2019
Mechanical Properties of Fluids
142790
Drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\rho_{0}$. If surface tension of liquid is $\mathrm{s}$, the radius of drop is
